# NCERT Solutions for Class 12 Physics Chapter 15 Communication System

## NCERT Solutions for Class 12 Physics Chapter 15 – Free PDF Download

*According to the latest term-wise CBSE Syllabus 2021-22, this chapter has been removed.

The NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems provided here are written by subject experts after extensive research on each and every topic to produce apt and authentic information for the students. If you go through the previous year question papers you will get a clear idea of questions directly asked from the book, in the examination. Hence, referring to these NCERT Solutions for Class 12 Physics during your preparations is a wise decision.

This NCERT Solutions Class 12 Physics presents different varieties of questions like MCQs, fill in the blanks, match the following, true or false, short answer questions along with numerical problems, important formulas, exercises and assignments. The PDF of the NCERT Solutions for Class 12 Physics Chapter 15 helps you to understand the concepts clearly and helps you to retain knowledge for a longer period of time.

Q.1: Which of the following frequencies will be suitable for beyond-the horizon communication using sky waves?

(1) 10 kHz

(2) 10 MHz

(3) 1 GHz

(4)  1000 GHz

Soln:

(2) 10 MHz

The signal waves need to travel a large distance for beyond – the – horizon communication.

Due to the antenna size, the 10 kHz signals cannot be radiated efficiently.

The 1 GHz – 1000 GHz (high energy) signal waves penetrate the ionosphere.

The 10 MHz frequencies get reflected easily from the ionosphere. Therefore, for beyond – the – horizon communication signal waves of 10 MHz frequencies are suitable.

Q.2: Frequencies in the UHF range normally propagate by means of :

(1) Ground Waves

(2) Sky Waves

(3) Surface Waves

(4)  Space Waves

Soln:

(4) Space Waves

Due to its high frequency, an ultra-high frequency (UHF) wave cannot travel along the trajectory of the ground also it cannot get reflected by the ionosphere. The ultrahigh-frequency signals are propagated through line – of – sight communication, which is actually space wave propagation.

Q.3: Digital signals

(i) Do not provide a continuous set of values

(ii) Represent value as discrete steps

(iii) Can utilize binary system

(iv) Can utilize decimal as well as binary systems

State which statement(s) are true?

(a) (1), (2) and (3)

(b) (1) and (2) only

(c) All statements are true

(d) (2) and (3) only

Soln:

(a) (1), (2) and (3) For transferring message signals the digital signals use the binary (0 and 1) system. Such a system cannot utilise the decimal system. Discontinuous values are represented in digital signals.

Q.4: Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?

Soln: In line – of – sight communication, between the transmitter and the receiver there is no physical obstruction. So, there is no need for the transmitting and receiving antenna to be at the same height.

Height of the antenna, h = 81 m

Radius of earth, R = 6.4 x 106m

d = √2Rh, for range

The service area of the antenna is given by the relation :

A = πd2 = π(2Rh)

= 3.14 x 2 x 6.4 x 106 x 81

= 3255.55 x 106 m2 = 3255.55 = 3256 km2

Q.5: A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?

Soln:

Given:

Amplitude of carrier wave, Ac = 12 V

Modulation index, m = 75% = 0.75

Amplitude of the modulating wave = Am

Modulation index is given by the relation :

m = $\frac{A_{m}}{A_{c}}$

Therefore, Am = m.Ac

= 0.75 x 12 V= 9 V

Q.6: A modulating signal is a square wave, as shown in the figure.

The carrier wave is given by c(t)=2sin(8πt)volts.

(1) Sketch the amplitude modulated waveform

(2) What is the modulation index?

Soln:

The amplitude of the modulating signal, Am = 1v can be easily observed from the given modulating signal.

Carrier wave is given by, c(t) = 2 sin(8nt)

Amplitude of the carrier wave, Ac = 2v

Time period, Tm = 1s

The angular frequency of the modulating signal is given by,

$\omega _{m} = \frac{2\pi }{T_{m}}$

The angular frequency of carrier signal, $\omega _{c} = 8\pi$ rad s-1         …(2)

from eqns.(1) and (2),

we get,   $\omega _{c} = 4\omega _{m}$

The modulating signal having the amplitude modulated waveform is shown in the figure:

(2) Modulation index, m = $\frac{A_{m}}{A_{c}}$ = $\frac{1}{2}$ = 0.5

Q.7: For an amplitude modulated wave, the maximum amplitude is found to be 10V while the minimum amplitude is found to be 2V. Determine the modulation index, µ. What would be the value of µ if the minimum amplitude is zero volts?

Soln:

Given,

Maximum Amplitude, Amax = 10 V

Minimum Amplitude, Amin = 2 V

For a wave, modulation index µ, is given by :

µ = $\frac{A_{max} – A_{min}}{A_{max} + A_{min}}$

= $\frac{10 – 2}{10 + 2}$ = $\frac{8}{12}$ = 0.67

If Amin = 0,

Then,

µ’ = $\frac{A_{max}}{ A_{max}}$ = 10/10 = 1

Q.8: Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.

Soln:  Let, $\omega _{c}$ be the carrier wave frequency

$\omega _{s}$ be the signal wave frequency

Signal received, V = V1 cos ($\omega _{c}$ + $\omega _{s}$)t

Instantaneous voltage of the carrier wave, Vm = Vc cos $\omega _{c}$t

V.Vin = V1cos($\omega _{c}$ + $\omega _{s}$)t. (Vc cos $\omega _{c}$t)

= V1Vc [cos($\omega _{c}$ + $\omega _{s}$)t . cos $\omega _{c}$t]

= $\frac{V_{1} V_{c}}{ 2} \left [ cos{( \omega _{c} + \omega _{s}) t + \omega _{c}t} + cos{( \omega _{c} + \omega _{s})t – \omega _{c}t} \right ]$

= $\frac{V_{1} V_{c}}{ 2} \left [ cos{( 2\omega _{c} + \omega _{s}) t + cos\omega _{s}t} \right ]$

The low pass filter allows only the high frequency signals to pass through it. The low frequency signal $\omega _{s}$ is obstructed by it.

Thus, at the receiving station, we can record the modulating signal, $\frac{V_{1}V_{C}}{2}cos\omega _{s}t$ which is the signal frequency.

• The transmitter, transmission channel, and receiver are three basic units of a communication system.
• Low frequencies cannot be transmitted to long distances. Therefore, they are superimposed on a high-frequency carrier signal by a process known as modulation.
• Two important forms of the communication system are Analog and Digital.
• Amplitude modulated waves can be produced by application of the message signal and the carrier wave to a non-linear device, followed by a bandpass filter.

## Class 12 Physics NCERT Solutions for Chapter 15 Communication Systems

The NCERT Solutions for Class 12 Physics will help you understand that communication systems are simply a collection of systems used for transmission, connection, communication and interconnection. These systems are categorically arranged into three different types on the basis of uses such as the Media, Technology and Application area. Sensors, Transducers, Emitters and Amplifiers are all examples of modern technology that are used as components in the majority of modern devices.

1. E-mail
2. Internet
3. Television
5. Computer

## Concepts involved in NCERT Class 12 Chapter 15 Communication System are:

1. Introduction
2. Elements Of A Communication System
3. Basic Terminology Used In Electronic Communication Systems
4. Bandwidth Of Signals
5. Bandwidth Of Transmission Medium
6. Propagation Of Electromagnetic Waves
1. Ground wave
2. Skywaves
3. Space wave
7. Modulation And Its Necessity
8. Size of the antenna or aerial
9. Effective power radiated by an antenna Ex 15.7.3 – Mixing up of signals from different transmitters
10. Amplitude Modulation
11. Production Of Amplitude Modulated Wave
12. Detection Of Amplitude Modulated Wave

To obtain a firm grip over these key concepts of this chapter and subject, students can access the NCERT Solutions that are developed by educational experts in the field. These solutions you to analyze your strengths and weaknesses thus providing a clear strategy to improve your academic performance.

In order to score good marks, it is very important for the students to also solve and get well versed with the NCERT exemplary questions and problems.

 Also Access NCERT Exemplar for Class 12 Physics Chapter 15 CBSE Notes for Class 12 Physics Chapter 15

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## Frequently Asked Questions on NCERT Solutions for Class 12 Physics Chapter 15

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